8) A particle moves from rest in a straight line with uniform acceleration. it covered 24 m . in the first four seconds of starting its motion, then the magnitude of its acceleration \( =\ldots \ldots \ldots . . \mathrm{m} \cdot / \mathrm{sec}^{2} \) \( \begin{array}{llll}\text { a) } \frac{3}{2} & \text { b) } 3 & \text { c) } 6 & \text { d) } 12\end{array} \)

To solve for the acceleration of the particle, we can use the equation of motion for uniformly accelerated linear motion: \[ s = ut + \frac{1}{2} a t^2 \] Here, \( s \) is the distance covered (24 m), \( u \) is the initial velocity (0 m/s, since it starts from rest), \( t \) is the time (4 s), and \( a \) is the acceleration we want to find. Plugging in the values: \[ 24 = 0 \cdot 4 + \frac{1}{2} a (4^2) \] This simplifies to: \[ 24 = \frac{1}{2} a \cdot 16 \] \[ 24 = 8a \] Now, solving for \( a \): \[ a = \frac{24}{8} = 3 \, \text{m/s}^2 \] So, the answer to the question is \( b) \, 3 \, \text{m/s}^2 \). This means the particle accelerates at a magnitude of 3 m/s²!